Divide $$$x^{3}$$$ by $$$x - 3$$$

Write the problem in the special format (missed terms are written with zero coefficients):

$$$\begin{array}{r|r}\hline\\x-3&x^{3}+0 x^{2}+0 x+0\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{x^{3}}{x} = x^{2}$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$x^{2} \left(x-3\right) = x^{3}- 3 x^{2}$$$.

Subtract the dividend from the obtained result: $$$\left(x^{3}\right) - \left(x^{3}- 3 x^{2}\right) = 3 x^{2}$$$.

$$\begin{array}{r|rrrr:c}&{\color{Fuchsia}x^{2}}&&&&\\\hline\\{\color{Magenta}x}-3&{\color{Fuchsia}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Fuchsia}x^{3}}}{{\color{Magenta}x}} = {\color{Fuchsia}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 3 x^{2}&&&{\color{Fuchsia}x^{2}} \left(x-3\right) = x^{3}- 3 x^{2}\\\hline\\&&3 x^{2}&+0 x&+0&\end{array}$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{3 x^{2}}{x} = 3 x$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$3 x \left(x-3\right) = 3 x^{2}- 9 x$$$.

Subtract the remainder from the obtained result: $$$\left(3 x^{2}\right) - \left(3 x^{2}- 9 x\right) = 9 x$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&{\color{Red}+3 x}&&&\\\hline\\{\color{Magenta}x}-3&x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 3 x^{2}&&&\\\hline\\&&{\color{Red}3 x^{2}}&+0 x&+0&\frac{{\color{Red}3 x^{2}}}{{\color{Magenta}x}} = {\color{Red}3 x}\\&&-\phantom{3 x^{2}}&&&\\&&3 x^{2}&- 9 x&&{\color{Red}3 x} \left(x-3\right) = 3 x^{2}- 9 x\\\hline\\&&&9 x&+0&\end{array}$$

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{9 x}{x} = 9$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$9 \left(x-3\right) = 9 x-27$$$.

Subtract the remainder from the obtained result: $$$\left(9 x\right) - \left(9 x-27\right) = 27$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&+3 x&{\color{Purple}+9}&&\\\hline\\{\color{Magenta}x}-3&x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 3 x^{2}&&&\\\hline\\&&3 x^{2}&+0 x&+0&\\&&-\phantom{3 x^{2}}&&&\\&&3 x^{2}&- 9 x&&\\\hline\\&&&{\color{Purple}9 x}&+0&\frac{{\color{Purple}9 x}}{{\color{Magenta}x}} = {\color{Purple}9}\\&&&-\phantom{9 x}&&\\&&&9 x&-27&{\color{Purple}9} \left(x-3\right) = 9 x-27\\\hline\\&&&&27&\end{array}$$

Since the degree of the remainder is less than the degree of the divisor, we are done.

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Fuchsia}x^{2}}&{\color{Red}+3 x}&{\color{Purple}+9}&&\text{Hints}\\\hline\\{\color{Magenta}x}-3&{\color{Fuchsia}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Fuchsia}x^{3}}}{{\color{Magenta}x}} = {\color{Fuchsia}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 3 x^{2}&&&{\color{Fuchsia}x^{2}} \left(x-3\right) = x^{3}- 3 x^{2}\\\hline\\&&{\color{Red}3 x^{2}}&+0 x&+0&\frac{{\color{Red}3 x^{2}}}{{\color{Magenta}x}} = {\color{Red}3 x}\\&&-\phantom{3 x^{2}}&&&\\&&3 x^{2}&- 9 x&&{\color{Red}3 x} \left(x-3\right) = 3 x^{2}- 9 x\\\hline\\&&&{\color{Purple}9 x}&+0&\frac{{\color{Purple}9 x}}{{\color{Magenta}x}} = {\color{Purple}9}\\&&&-\phantom{9 x}&&\\&&&9 x&-27&{\color{Purple}9} \left(x-3\right) = 9 x-27\\\hline\\&&&&27&\end{array}$$

Therefore, $$$\frac{x^{3}}{x - 3} = \left(x^{2} + 3 x + 9\right) + \frac{27}{x - 3}$$$.